Question

Solve the equation \[\dfrac{x}{6} + 11 = \dfrac{1}{{15}}\] and check the result.

Answer 1

Hint:
In this question, we are required to find the solution of a linear equation. So, we will first arrange all the terms such that the like terms are together. We will then perform the arithmetic operations on the like terms as required. Comparing both sides, we will get our answer. Again, we will substitute the value obtained in the equation to check our solution.

Complete step by step solution:
Let us begin by identifying the like terms in our linear equation.
In the given linear equation, we have two types of terms, the variable terms, and the numerical terms. The variable term is \[\dfrac{x}{6}\] and the numerical terms are 11 and \[\dfrac{1}{{15}}\].
Now, let us arrange these terms together. We can arrange our variable terms on the LHS of the equation and we can arrange our numerical terms on the RHS of the equation. To do that, we will have to eliminate 11 from the LHS of the equation.
\[\dfrac{x}{6} + 11 = \dfrac{1}{{15}}\]
Subtracting 11 from both side, we get
\[\begin{array}{l} \Rightarrow \dfrac{x}{6} + 11 - 11 = \dfrac{1}{{15}} - 11\\ \Rightarrow \dfrac{x}{6} = \dfrac{1}{{15}} - 11\end{array}\]
We will now solve the like terms in our equation using the various arithmetic operations.
Let us begin by solving the RHS of the equation.
\[\dfrac{1}{{15}} - 11\]
Now, we will first take the LCM of the denominator. The LCM of 15 and 1 is 15.
We will now multiply the denominators with the suitable numbers to get the product as 15. Subsequently, we will multiply the numerators with similar numbers too, to not alter the fractions.
\[\begin{array}{l}\dfrac{1}{{15}} - 11 = \dfrac{{1 \times 1}}{{15 \times 1}} - \dfrac{{11 \times 15}}{{1 \times 15}}\\ = \dfrac{1}{{15}} - \dfrac{{165}}{{15}}\end{array}\]
Now, our denominators have been equalized so we can perform the arithmetic operation on our numerators. Hence, let us subtract the terms.
\[\dfrac{1}{{15}} - \dfrac{{165}}{{15}} = \dfrac{{ - 164}}{{15}}\]
As we have simplified our RHS, we can equate it to LHS now.
\[\dfrac{x}{6} = \dfrac{{ - 164}}{{15}}\]
Since we have to find the value of \[x\], so, we will now eliminate 6 from the LHS of the equation .
Multiplying both sides by 6, we get
\[\begin{array}{l}\dfrac{x}{6} \times 6 = \dfrac{{ - 164}}{{15}} \times 6\\ \Rightarrow x = \dfrac{{ - 984}}{{15}}\end{array}\]
Now we will divide the numerator with the denominator to obtain the value of \[x\].
\[x = - 65.6\]
We will now verify the solution, for that we will substitute the value of \[x\] obtained into the original linear equation. If the equation is satisfied, i.e., LHS becomes equal to the RHS, then our solution will be correct, otherwise, it will be false.
Substitute \[ - 65.6\] for \[x\] in the equation \[\dfrac{x}{6} + 11 = \dfrac{1}{{15}}\].
\[\dfrac{{ - 65.6}}{6} + 11 = \dfrac{1}{{15}}\]
Simplifying the equation, we get
\[\begin{array}{l} - 10.933 + 11 = 0.067\\0.067 = 0.067\end{array}\]

Since, LHS = RHS hence, \[x = - 65.6\] is verified.

Note:
The equation provided to us is a linear equation in a single variable. This means that the equation consists of only one unknown variable. The word linear implies that the degree of the equation is one. The degree of an equation is referred to as the highest power of the variable. Since the equation is linear in one variable, only one equation is required to solve it.